NCERT Solutions for Class 12 Science Physics Chapter 8 - Electromagnetic Waves

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NCERT Solutions for Class 12 Science Physics Chapter 8 Electromagnetic Waves are provided here with simple step-by-step explanations. These solutions for Electromagnetic Waves are extremely popular among Class 12 Science students for Physics Electromagnetic Waves Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Class 12 Science Physics Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class Class 12 Science Physics are prepared by experts and are 100% accurate.

Page No 47:

Question 1:

One requires 11eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in

(a) visible region.
(b) infrared region.
(c) ultraviolet region.
(d) microwave region.

Answer:

Given: Energy required to dissociate a carbon monoxide molecule into carbon and oxygen atoms, E = 11 eV
Energy, $E = h ν$
Here h is Planck's constant $= 6.62 \times 10^{- 34} J - s$
$\Rightarrow 11 \times 1.6 \times 10^{- 19} = 6.62 \times 10^{- 34} ν \Rightarrow ν = 2.65 \times 10^{15} Hz$
This frequency radiation is under the ultraviolet region.
Hence, the correct answer is option C.

Page No 47:

Question 2:

A linearly polarized electromagnetic wave given as $E = E_{0} \overset{\land}{i} \cos (k z - ω t)$ is incident normally on a perfectly reflecting
infinite wall at z = a. Assuming that the material of the wall is optically inactive, the reflected wave will be given as

$(a) E_{r} = - E_{0} \hat{i} \cos (k z - ω t) . (b) E_{r} = E_{0} \hat{i} \cos (k z + ω t) . (c) E_{r} = - E_{0} \hat{i} \cos (k z + ω t) . (d) E_{r} = E_{0} \hat{i} \sin (k z - ω t) .$

Answer:

Given: Incident electromagnetic wave, $E = E_{0} \overset{\land}{i} \cos (k z - ω t)$
When a wave is reflected from a denser medium, then the phase changes by 180° or π radian.
Reflected electromagnetic wave is given by:
$\begin{array}{rcl} E_{r} & = & E_{0} (- \overset{\land}{i}) \cos [k (- z) - ω t + π] \\ = & - E_{0} \overset{\land}{i} \cos [- (k z + ω t) + π] \\ = & E_{0} \overset{\land}{i} \cos (- (k z + ω t)) ∵ \cos (π + θ) = - \cos θ \\ = & E_{0} \overset{\land}{i} \cos (k z + ω t) ∵ \cos (- θ) = \cos θ \end{array}$
Hence, the correct answer is option B.

Page No 48:

Question 3:

Light with an energy flux of 20 W/cm² falls on a non-reflecting surface at normal incidence. If the surface has an area of 30 cm². the total momentum delivered (for complete absorption) during 30 minutes is

(a) 36 × 10^–5kg m/s.
(b) 36 × 10^–4 kg m/s.
(c) 108 × 10⁴ kg m/s.
(d) 1.08 × 10⁷ kg m/s.

Answer:

Given: Intensity of incident light, I = 20 W/cm²
Surface area, A = 30 cm²
Time, t = 30 minutes
Total energy falling on the surface in time t, E = IAt
$\Rightarrow E = 20 \times 30 \times 30 \times 60 J$
Momentum of incident light:
$p_{i} = \frac{E}{c} = \frac{20 \times 30 \times 30 \times 60}{3 \times 10^{8}} = 36 \times 10^{- 4} kgm / s$
Momentum of reflected light:
$p_{f} = 0$
Momentum delivered to the surface for complete absorption is $36 \times 10^{- 4} kgm / s$ .
Hence, the correct answer is option B.

Page No 48:

Question 4:

The electric field intensity produced by the radiations coming from 100 W bulb at a 3 m distance is E. The electric field intensity produced by the radiations coming from 50 W bulb at the same distance is

$(a) \frac{E}{2} (b) 2 E (c) \frac{E}{\sqrt{2}} (d) \sqrt{2} E$

Answer:

Given: Power of first bulb, P₁ = 100 W
Power of second bulb, P₂ = 50 W
Distance, d = 3 m
Electric field intensity:
$E \propto P \Rightarrow \frac{E_{1}}{E_{2}} = \frac{P_{1}}{P_{2}} = \frac{100}{50} = 2 \Rightarrow E_{2} = \frac{E_{1}}{2}$
So, the electric field intensity produced by the radiations coming from a 50 W bulb is $\frac{E_{1}}{2} = \frac{E}{2}$ .
Hence, the correct answer is option A.

Page No 48:

Question 5:

If E and B represent electric and magnetic field vectors of the electromagnetic wave, the direction of propagation of electromagnetic wave is along
(a) E.
(b) B.
(c) B × E.
(d) E × B.

Answer:

The direction of propagation of the electromagnetic wave is perpendicular to both electric field vector E and magnetic field vector B.

Here, an electromagnetic wave is propagating along the z-direction which is given by the cross product of E and B.
Hence, the correct answer is option D.

Page No 48:

Question 6:

The ratio of contributions made by the electric field and magnetic field components to the intensity of an EM wave is
(a) c : 1
(b) c²: 1
(c) 1 : 1
(d) $\sqrt{c} : 1$

Answer:

Intensity due to electric field, $I_{E} = \frac{1}{2} c ε_{0} E^{2} . . . . . (1)$
Intensity due to magnetic field, $I_{B} = \frac{c B^{2}}{2 μ_{0}} . . . . . (2)$
Speed of light, $c = \frac{1}{\sqrt{μ_{0} ε_{0}}} and c = \frac{E}{B}$
Using equation (1) and (2),

$\frac{I_{E}}{I_{B}} = \frac{\frac{1}{2} c ε_{0} E^{2}}{\frac{1}{2} \frac{c B^{2}}{μ_{0}}} = \frac{ε_{0} E^{2}}{\frac{B^{2}}{μ_{0}}} = (ε_{0} μ_{0}) {(\frac{E}{B})}^{2}$
$\frac{I_{E}}{I_{B}} = \frac{1}{c^{2}} c^{2} = \frac{1}{1}$
So, the ratio of contribution made by electric and magnetic field components to the intensity of an EM wave is 1 : 1
Hence, the correct answer is option C.

Page No 48:

Question 7:

An EM wave radiates outwards from a dipole antenna, with E₀as the amplitude of its electric field vector. The electric field E₀ which transports significant energy from the source falls off as

$(a) \frac{1}{r^{3}} (b) \frac{1}{r^{2}} (c) \frac{1}{r}$
(d) remains constant.

Answer:

Given: Amplitude of electric field vector = E_o
Intensity of radiation, $I = \frac{W}{A t}$
Here, W is energy emitted by source t is time A is area on which radiation is incident, If energy and time are constant
$\Rightarrow I \propto \frac{1}{A}$
An antenna radiates energy in spherical shape
$\Rightarrow I \propto \frac{1}{r^{2}} [∵ A = 4 π r^{2}]$
$Intensity, I = \frac{1}{2} ε_{o} {E_{o}}^{2} \Rightarrow {E_{o}}^{2} \propto \frac{1}{r^{2}} \Rightarrow E_{o} \propto \frac{1}{r}$
Hence, the correct answer is option C.

Page No 49:

Question 8:

An electromagnetic wave travels in vacuum along z-direction: $E = (E_{1} \hat{i} + E_{2} \hat{j}) \cos (k z - ω t)$ Choose the correct options from the following:

(a) The associated magnetic field is given as $B = \frac{1}{c} (E_{1} \hat{i} + E_{2} \hat{j}) \cos (k z - ω t)$
(b) The associated magnetic field is given as $B = \frac{1}{c} (E_{1} \hat{i} - E_{2} \hat{j}) \cos (k z - ω t)$
(c) The given electromagnetic field is circularly polarised.
(d) The given electromagnetic wave is plane polarised.

Answer:

Given: Electric field vector, $E = (E_{1} \hat{i} + E_{2} \hat{j}) \cos (k z - ω t)$
Relation between electric field and magnetic field:-
$B = \frac{E}{c} = \frac{1}{c} (E_{1} \hat{i} + E_{2} \hat{j}) \cos (k z - ω t)$
Em wave propagates through z-direction and E and B vectors lie in x-y plane. Here, the propagation of electromagnetic wave is perpendicular to E and B. This means electromagnetic wave is plane polarised.
Hence, the correct answer is option A and D.

Disclaimer: option a and b were same in question, so option a is corrected.

Page No 49:

Question 9:

An electromagnetic wave travelling along z-axis is given as: E = E₀ cos (kz − ωt.). Choose the correct options from the following;

(a) The associated magnetic field is given as $B = \frac{1}{c} \hat{k} \times E = \frac{1}{ω} (k \times E)$
(b) The electromagnetic field can be written in terms of the associated magnetic field as E =c $(B \times \hat{k})$
(c) $\hat{k} . E = 0, \hat{k} . B = 0 .$
(d) $\hat{k} \times E = 0, k \times B = 0 .$

Answer:

Given: Electric field, E = E₀ cos (kz − ωt.).
Associated magnetic field, $B_{0} = \frac{E_{0}}{c}$
Also, this magnetic field is along x-axis in an electromagnetic wave
i.e., $\vec{B} = \frac{1}{c} (\hat{k} \times \vec{E})$
$\vec{B} = \frac{1}{ω} (\vec{k} \times \vec{E}) (∵ c = k ω)$
$Similarly, Electric field, \vec{E} = c (\vec{B} \times \hat{k})$
Since direction of $\hat{k}$ is perpendicular to both $\vec{E}$ and $\vec{B}$ , so
$\hat{k} . \vec{E} = 0, \hat{k} . \vec{B} = 0$
Hence, the correct answer is option a, b and c.

Disclaimer: Option a is corrected, it should be $B = \frac{1}{c} \hat{k} \times E = \frac{1}{ω} (k \times E)$ instead of $B = \frac{1}{c} \hat{k} \times E = \frac{1}{ω} (\hat{k} \times E)$ .

Page No 49:

Question 10:

A plane electromagnetic wave propagating along x direction can have the following pairs of E and B
(a) E_x, B_y.
(b) E_y, B_z.
(c) B_x, E_y.
(d) E_z, B_y

Answer:

Electric field and magnetic field vectors are perpendicular to each other and the direction of propagation of electromagnetic wave.

The following pairs of E and B can be E_y, B_zor E_z, B_y.
Hence the correct answer is option (B, D).

Page No 50:

Question 11:

A charged particle oscillates about its mean equilibrium position with a frequency of 10⁹ Hz. The electromagnetic waves produced:

(a) will have frequency of 10⁹ Hz.
(b) will have frequency of 2 × 10⁹ Hz.
(c) will have a wavelength of 0.3 m.
(d) fall in the region of radiowaves.

Answer:

The frequency of em wave is equal to the frequency of oscillation of charged particle producing em wave.
Frequency, $ν = 10^{9} Hz$ (lies in radio waves)
Wavelength, $λ = \frac{c}{ν}$
Here, c = speed of light = 3 × 10⁸ m/s
$\Rightarrow λ = \frac{3 \times 10^{8}}{10^{9}} = 0.3 m$
Hence, the correct answer is option (A, C, D).

Page No 50:

Question 12:

The source of electromagnetic waves can be a charge

(a) moving with a constant velocity.
(b) moving in a circular orbit.
(c) at rest.
(d) falling in an electric field.

Answer:

An accelerated charge can emit em waves. When a charged particle is in motion inside an electric field then its velocity keeps on changing along with its acceleration.
So, a charge moving in a circular orbit or falling in an electric field has acceleration.
Hence, the correct answer is option (B, D).

Page No 50:

Question 13:

An EM wave of intensity I falls on a surface kept in vacuum and exerts radiation pressure p on it. Which of the following are true?

(a) Radiation pressure is I/c if the wave is totally absorbed.
(b) Radiation pressure is I/c if the wave is totally reflected.
(c) Radiation pressure is 2I/c if the wave is totally reflected.
(b) Radiation pressure is in the range I/c < p < 2I/c for real surfaces.

Answer:

Change in momentum per unit time is given as:
$P = \frac{Intensity (I)}{Speed of wave (c)} = Radiation pressure$
If wave is completely absorbed by surface then the momentum of reflected wave per unit area is zero.
So, radiation pressure when wave is absorbed is $\frac{∆ I}{c} = \frac{I}{c} - 0 = \frac{I}{c}$
Now, when the wave is entirely reflected then $p = \frac{I}{c} - (- \frac{I}{c}) = \frac{2 I}{c}$
If the surface is not perfectly absorbing and perfectly reflecting, then p will be more than $\frac{I}{c}$ but less than $\frac{2 I}{c}$
Hence, the correct answer is option A, C and D.

Page No 50:

Question 14:

Why is the orientation of the portable radio with respect to broadcasting station important?

Answer:

The orientation of the portable radio with respect to broadcasting station is important this is because for receiving electromagnetic waves receiving antenna should be parallel to the vibration of the electric or magnetic field of the wave.

Page No 50:

Question 15:

Why does microwave oven heats up a food item containing water molecules most efficiently?

Answer:

Microwave oven heats up a food item containing water molecules most efficiently because the frequency of microwave matches with the resonant frequency of water molecules. At resonant frequency, energy from the waves is transferred efficiently to the kinetic energy of water molecules.

Page No 50:

Question 16:

The charge on a parallel plate capacitor varies as q = q₀ cos 2πνt. The plates are very large and close together (area = A, separation = d). Neglecting the edge effects, find the displacement current through the capacitor?

Answer:

Given: Charge on a parallel plate capacitor, $q = q_{o} \cos 2 π ν t$
The displacement current is equal to the conduction current
$\Rightarrow I_{D} = I_{C} = \frac{d q}{d t} \Rightarrow I_{D} = I_{C} = \frac{d}{d t} (q_{o} \cos 2 π v t) = - 2 π v q_{o} \sin 2 π v t$
Hence, the required displacement current through the capacitor is $- 2 π v q_{o} \sin 2 π v t$ .

Page No 50:

Question 17:

A variable frequency a.c source is connected to a capacitor. How will the displacement current change with decrease in frequency?

Answer:

Capacitive reactance, X_c= $\frac{1}{ω C} = \frac{1}{2 π f C}$
From the above equation, the decrease in frequency of a.c. will increase the reactance of the capacitor and hence will decrease the conduction current.
Also, in this case the displacement current is equal to conduction current, so displacement current will decrease with the decrease in frequency of a.c.

Page No 50:

Question 18:

The magnetic field of a beam emerging from a filter facing a floodlight is given by B₀ = 12 × 10^–8sin (1.20 × 10⁷ z – 3.60 × 10¹⁵ t) T. What is the average intensity of the beam?

Answer:

Given magnetic feild, $B_{o} = 12 \times 10^{- 8} \sin (1.20 \times 10^{7} z - 3.60 \times 10^{15} t) T$
General equation of magnetic field, B = B_o sin ωt
On comparing the general equation and the given equation, we get
$B_{o} = 12 \times 10^{- 8}$
Average intensity, $I_{a v} = \frac{1}{2} \frac{{B_{o}}^{2} c}{μ_{o}}$
$\Rightarrow I_{a v} = \frac{1}{2} \times \frac{{(12 \times 10^{- 8})}^{2} (3 \times 10^{8})}{4 π \times 10^{- 7}} \Rightarrow I_{a v} = 1.71 W / m^{2}$
Hence, the required average intensity is 1.71 W/m².

Page No 50:

Question 19:

Poynting vectors S is defined as a vector whose magnitude is equal to the wave intensity and whose direction is along the direction of wave propogation. Mathematically, it is given by $S = \frac{1}{μ_{0}} E \times B$ Show the nature of S vs t graph.

Answer:

Let us consider an electromagnetic wave in which electric field $\vec{E}$ is varying along y-axis, magnetic field $\vec{B}$ is varying along Z-axis and the propagation of wave is along x-axis.
Let $\vec{E} = E_{0} \sin (ω t - k x) \hat{j}$
$\vec{B} = B_{0} \sin (ω t - k x) \hat{k}$
Poynting vector, $S = \frac{1}{μ_{0}} (\vec{E} \times \vec{B})$
$\Rightarrow S = \frac{1}{μ_{0}} E_{0} B_{0} \sin^{2} (ω t - k x) \hat{i}$
Since $\sin^{2} (ω t - k x)$ is never negative and $\vec{s} (x, t)$ always points in the positive x-direction. So, the variation of |S| w.r.t. t can be shown as:

Page No 51:

Question 20:

Professor C.V Raman surprised his students by suspending freely a tiny light ball in a transparent vacuum chamber by shining a laser beam on it. Which property of EM waves was he exhibiting? Give one more example of this property.

Answer:

Electromagnetic waves carry energy and momentum like other waves. By exerting momentum, pressure will be exerted known as radiation pressure. The tail of comet is also due to this property. Professor CV Raman was surprised by shining beam on it which proves the propagation of em waves from vacuum.

Page No 51:

Question 21:

Show that the magnetic field B at a point in between the plates of a parallel-plate capacitor during charging is $\frac{\in_{0} μ_{0} r}{2} \frac{d E}{d t}$ (symbols having usual meaning).

Answer:

Let us consider parallel plates capacitor having plates of radius r.

Magnetic field at a point in between the plates of the parallel plate capacitor, if I_Dis the displacement current.
$B = \frac{μ_{0}}{4 π} \frac{2 I_{D}}{r} = \frac{μ_{0}}{2 π} \frac{I_{D}}{r}$
As $I_{D} = \in_{0} \frac{d ϕ_{E}}{d t}$
$\Rightarrow B = \frac{μ_{0}}{2 π} \frac{\in_{0}}{r} \frac{d ϕ_{E}}{d t} . . . . . (1)$
If E is the electric field between the plates of capacitor, then
$ϕ_{E} = E \times π r^{2} [∵ ϕ_{E} = E \times A]$
$\Rightarrow \frac{d ϕ_{E}}{d t} = π r^{2} \frac{d E}{d t} . . . . . (2)$
Using equation (1) and (2)
$B = \frac{μ_{0} \in_{0}}{2 π r} \times π r^{2} \frac{d E}{d t} = \frac{\in_{0} μ_{0} r}{2} \frac{d E}{d t}$
Hence Proved.

Disclaimer : In question $\frac{\in_{0} μ_{r}}{2} \frac{d E}{d t}$ is replaced by $\frac{\in_{0} μ_{0} r}{2} \frac{d E}{d t}$

Page No 51:

Question 22:

Electromagnetic waves with wavelength

(i) λ₁ is used in satellite communication.
(ii) λ₂ is used to kill germs in water purifies.
(iii) λ₃ is used to detect leakage of oil in underground pipelines.
(iv) λ₄ is used to improve visibility in runways during fog and mist conditions.
(a) Identify and name the part of electromagnetic spectrum to which these radiations belong.
(b) Arrange these wavelengths in ascending order of their magnitude.
(c) Write one more application of each.

Answer:

(a)
(i) For satellite communication $\to$ Microwaves
(ii) For killing germs $\to$ Ultraviolet rays
(iii) For detecting leakage of oil $\to$ x - rays
(iv) For improving visibility $\to$ Infrared

(b) Ascending order is
$λ_{3} < λ_{2} < λ_{4} < λ_{1}$
Microwaves - 1 mm – 25 μm
UV rays - 400 nm – 1 nm
X-rays - 1 nm – 1 pm
Infrared - 25 μm – 2.5 μm

(c)
(i) Microwaves are used in radar system.
(ii) UV rays are used in Lasik surgery
(iii) X-rays is used for treating bones
(iv) Infrared in optical communication.

Page No 51:

Question 23:

Show that average value of radiant flux density ‘S’ over a single period ‘T’ is given by $S = \frac{1}{2 c μ_{0}} {E_{0}}^{2}$

Answer:

Let us consider electromagnetic waves be propagating along x-axis if electric field is along y-axis and magnetic field vector is along z-axis. Then
E = E₀ cos (kx – ωt)
B = B₀ cos (kx – ωt)

Radiant flux density, $S = \frac{1}{μ_{0}} (E \times B)$
= c²∈₀(E × B)
S = c²∈₀(E₀ × B₀) cos²(kx – ωt)
Average value of radiant flux density,
$S_{av} = c^{2} \in_{0} | E_{0} \times B_{0} | \frac{1}{T} \int_{0}^{T} \cos^{2} (k x - ω t) d t$
$\Rightarrow S_{av} = \frac{c^{2}}{2} \in_{0} E_{0} (\frac{E_{0}}{c}) \Rightarrow S_{av} = \in_{0} \frac{E_{0}^{2} c}{2} = \frac{c}{2} \times \frac{1}{c^{2} μ_{0}} E_{0}^{2} \Rightarrow S_{av} = \frac{E_{0}^{2}}{2 μ_{0} c}$

Page No 51:

Question 24:

You are given a 2μF parallel plate capacitor. How would you establish an instantaneous displacement current of 1mA in the space between its plates?

Answer:

Given: Displacement current, I_D = 1 mA
Capacitance C = 2μF
Charge, Q = cv and charge, Q = I_Ddt
$\Rightarrow I_{D} = C \frac{d v}{d t} \Rightarrow \frac{d v}{d t} = \frac{I_{D}}{C} = \frac{10^{- 3}}{2 \times 10^{- 6}} = 500 v / s$
Hence, by applying a varying potential difference of 500 v/s, a displacement current of desired value could be produced.

Page No 51:

Question 25:

Show that the radiation pressure exerted by an EM wave of intensity I on a surface kept in vacuum is I/c.

Answer:

Let us consider intensity falls on the surface of area A be I.
Energy received by surface per second = I × A
Let N be the number of photons per second.

$\Rightarrow N = \frac{E}{E_{photon}} = \frac{E}{\frac{h c}{λ}} \Rightarrow N = \frac{I A λ}{h c}$
If surface is perfectly absorbing:
$∆ P = \frac{h}{λ}$
Force, F = N × $∆ P = \frac{I A}{c}$
Pressure = $\frac{Force}{Area} = \frac{I A}{c A} = \frac{I}{c}$
This is the required relation.

Page No 51:

Question 26:

What happens to the intensity of light from a bulb if the distance from the bulb is doubled? As a laser beam travels across the length of a room, its intensity essentially remains constant.

What geomatrical characteristic of LASER beam is responsible for the constant intensity which is missing in the case of light from the bulb?

Answer:

Given: Distance from the bulb is doubled.
Intensity of light, $I \propto \frac{1}{r^{2}} [∵ A = 4 π r^{2}]$
This means intensity of light reduces to one-fourth. But the laser beam does not spread as it travels across the length of a room, so its intensity remains constant.

Some geometrical characteristics of laser beam which are missing in case of normal light from the bulb are:
(i) Unidirectional
(ii) Monochromatic
(iii) Coherent light
(iv) Highly collimated.

Page No 52:

Question 27:

Even though an electric field E exerts a force qE on a charged particle yet the electric field of an EM wave does not contribute to the radiation pressure (but transfers energy). Explain.

Answer:

The electric field of an electromagnetic wave is an oscillating field and so is the electric force caused by it on a charged particle. This electric force averaged over an integral number of cycles is zero since its direction changes every half cycle.
Therefore, electric field is not responsible for radiation pressure.

Page No 52:

Question 28:

An infinitely long thin wire carrying a uniform linear static charge density λ is placed along the z-axis (Fig. 8.1). The wire is set into
motion along its length with a uniform velocity $v = ν {\hat{k}}_{z}$ . Calculate the poynting vector $S = \frac{1}{μ_{0}} (E \times B)$

Answer:

Let us consider a cylindrical Gaussian surface in such a way that the axis of cylinder lies on wire. Electric field intensity due to long straight wire at a distance r and charge density $λ \frac{c}{m} .$

Electric field due to infinitely long thin wire, $\vec{E} = \frac{λ}{2 π \in_{0} r} \hat{j}$
Magnetic field due to the wire, $\vec{B} = \frac{μ_{0} I}{2 π r} \hat{i}$
Equivalent current flowing through the wire:-
$I = \frac{Q}{t} = \frac{λ l}{t} = λ ν \Rightarrow \vec{B} = \frac{μ_{0} λ ν}{2 π r} \hat{i}$
Poynting vector, $S = \frac{1}{μ_{0}} [\vec{E} \times \vec{B}]$
$= \frac{1}{μ_{0}} [\frac{λ}{2 π \in_{0} r} \hat{j} \times \frac{μ_{0} λ ν}{2 π r} \hat{i}] = \frac{λ^{2} ν}{4 π^{2} r^{2} \in_{0}} (\hat{j} \times \hat{i}) = \frac{- λ^{2} ν}{4 π^{2} r^{2} \in_{0}} \hat{k}$

Hence, this is the required value of poynting vector.

Page No 52:

Question 29:

Sea water at frequency ν = 4 × 10⁸ Hz has permittivity ε ≈ 80 ε_o, permeability μ ≈ μ_o and resistivity ρ = 0.25 Ω–m. Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source V(t) = V_o sin (2π νt). What fraction of the conduction current density is the displacement current density?

Answer:

Given: Frequency, $ν = f = 4 \times 10^{8} Hz$
Permittivity, $\in \approx 80 \in_{0}$
Permeability, $μ = μ_{0}$
Resistivity, ρ = 0.25Ω m
Let d be the distance between the plates of the parallel plate capacitor.
Electric field between the plates, $E = \frac{v}{d} (t)$
$\Rightarrow E = \frac{v_{0}}{d} \sin (2 π ν t)$
Using ohm's law, conduction current density is given by
$J^{C} = \frac{E}{ρ} = \frac{v_{0}}{ρ d} \sin (2 π ν t) \Rightarrow J^{C} = J_{0}^{C} \sin (2 π ν t)$
Displacement current density:-
$J^{d} = \in \frac{d E}{d t} = \in \frac{d}{d t} [\frac{V_{0}}{d} \sin (2 π v t)] = \frac{2 π v \in v_{0}}{d} \cos (2 π v t) \Rightarrow J^{d} = J_{0}^{d} \cos (2 π v t)$
This means $J_{0}^{C} = \frac{v_{0}}{ρ d}$ and $J_{0}^{d} = \frac{2 π v \in v_{0}}{d}$
$\Rightarrow \frac{J_{0}^{d}}{J_{0}^{d}} = \frac{2 π v \in ν_{0} \times ρ d}{d \times v_{0}} = 2 π ν \in ρ \Rightarrow \frac{J_{0}^{d}}{J_{0}^{c}} = \frac{2 π (4 \times 10^{8}) (80 \in_{0}) (0.25)}{1} \Rightarrow \frac{J_{0}^{d}}{J_{0}^{c}} = 4 π \in_{0} (4 \times 10^{9}) = \frac{4 \times 10^{9}}{9 \times 10^{9}} \Rightarrow \frac{J_{0}^{d}}{J_{0}^{c}} = \frac{4}{9}$

Page No 52:

Question 30:

A long straight cable of length l is placed symmetrically along z-axis and has radius a(<<l). The cable consists of a thin wire and
a co-axial conducting tube. An alternating current I(t) = I_o sin (2πνt) flows down the central thin wire and returns along the co-axial conducting tube. The induced electric field at a distance s from the wire inside the cable is $E (s, t) = μ_{0} I_{0} v \cos (2 πvt) l n (\frac{s}{a}) \hat{k} .$
(i) Calculate the displacement current density inside the cable.
(ii) Integrate the displacement current density across the cross-section of the cable to find the total displacement current I^d.
(iii) Compare the conduction current I₀with the dispalcement current $I_{0}^{d}$

Answer:

Given→ E(s, t) = µ₀I₀v cos $(2 π ν t) l n (\frac{s}{a}) \hat{k}$
= Displacement current density, J_d= $\in_{0} \frac{d E}{d t}$
$= \in_{0} \frac{d}{d t} [μ_{0} I_{0} ν \cos (2 π ν ι) l n (\frac{s}{a}) \hat{k}] = \in_{0} μ_{0} I_{0} ν \frac{d}{d t} [\cos (2 π ν t)] \ln (\frac{s}{a}) \hat{k} = \frac{1}{C^{2}} I_{0} ν^{2} 2 π [- \sin 2 π ν t] l n (\frac{s}{a}) \hat{k} = \frac{v^{2}}{C^{2}} I_{0} 2 π [- \sin 2 π ν t] l n (\frac{3}{a}) \hat{k} = \frac{1}{λ^{2}} 2 π I_{0} \sin 2 π ν t l n (\frac{a}{s}) \hat{k} = \frac{2 π I_{0}}{λ^{2}} l n (\frac{a}{s}) \sin 2 π ν t \hat{k}$

(ii) Total displacement current, $I_{d} = \int_{} J_{d} s d s \cos θ$
$= \int_{s = 0}^{a} J_{d} s d s \int_{0}^{2 π} d θ = \int_{s = 0}^{a} J_{d} s d s \times 2 π = \int_{0}^{a} [\frac{2 π I_{0}}{λ^{2}} l n (\frac{a}{s}) s d s \sin 2 π v t] 2 π = {(\frac{2 π}{λ})}^{2} I_{0} \int_{0}^{a} l n (\frac{a}{s}) s d s \sin 2 π v t = {(\frac{2 π}{λ})}^{2} I_{0} \int_{0}^{a} l n (\frac{a}{s}) \frac{1}{2} d {(s)}^{2} \sin 2 π v t = \frac{a^{2}}{2} {(\frac{2 π}{λ})}^{2} I_{0} \sin 2 π v t \int_{s = 0}^{a} l n (\frac{a}{s}) d {(\frac{s}{a})}^{2} = - \frac{a^{2}}{4} {(\frac{2 π}{λ})}^{2} I_{0} \sin 2 π v t \times (- 1) [∵ \int_{s = 0}^{a} l n (\frac{s}{a}) d {(\frac{s}{a})}^{2} = - 1]$
$\begin{array}{rcl} I_{D} & = & \frac{a^{2}}{4} {(\frac{2 π}{λ})}^{2} I_{0} \sin 2 π v t \\ = & {(\frac{2 π a}{2 λ})}^{2} I_{0} \sin 2 π v t \end{array}$
(iii) Displacement current = $\begin{array}{rcl} I_{d} & = & {(\frac{2 π a}{2 λ})}^{2} I_{0} \sin 2 π v t \end{array}$
$\begin{array}{rcl} = & I_{0}^{d} \sin 2 π v t \end{array}$
$Here, I_{0}^{d} = {(\frac{2 π a}{2 λ})}^{2} I_{0 =} {(\frac{a π}{λ})}^{2} I_{0} \Rightarrow \frac{I_{0}^{d}}{I_{0}} = {(\frac{a π}{λ})}^{2}$

Hence, the comparison between conduction current and displacement current is ${(\frac{a π}{λ})}^{2}$

Page No 52:

Question 31:

A plane EM wave travelling in vacuum along z direction is given by E = E₀ sin $(k z - ω t) \hat{i} a n d B = B_{0} \sin (k z - ω t) \hat{j}$

(i) Evaluate ∮E.dl over the rectangular loop 1234 shown in Fig 8.2.
(ii) Evaluate ∫B.ds over the surface bounded by loop 1234.
(iii) Use equation $\oint E . dl = \frac{- d ϕ_{B}}{d t} to prove \frac{E_{0}}{B_{0}} = c .$
(iv) By using similar process and the equation $\oint B . dl = μ_{0} I + \in_{0} \frac{d ϕ_{\in}}{d t}, prove that c = \frac{1}{\sqrt{μ_{0} \in_{0}}}$

Answer:

As shown in the diagram

(i) $\oint \vec{E} . d \vec{l} = \int_{1}^{2} \vec{E} . d \vec{l} + \int_{2}^{3} \vec{E} . d \vec{l} + \int_{3}^{4} \vec{E} . d \vec{l} + \int_{4}^{1} \vec{E} . d \vec{l}$
But angle between electric field
$\vec{E}$ = E₀ sin(kz – $ω$ t)i and side 1-2 and side 3-4 is 90º.
But angle between $\vec{E}$ and side 2-3 and 4-1 is 0º and 180º respectively
So, $\oint \vec{E} . d \vec{l} = 0 + \int_{2}^{3} E_{0} \sin (k z - ω t) d l \cos 0 º + 0 + \int_{4}^{1} E_{0} \sin (k z - ω t) d l \cos 180 º$
⇒ $\oint \vec{E} . d \vec{l} =$ E₀sin(kz₂– $ω$ t)h – E₀sin(kz₁– $ω$ t)h
= E₀h [sin(kz₂– $ω$ t) – sin(kz₁– $ω$ t)] .....(1)
(ii) Let us take a small element of area ds = h × dz as shown in the diagram

Given, $\vec{B} = B_{0} \sin (k z - ω t) \hat{j}$
$\begin{array}{l} So, \int \vec{B} . d \vec{s} = \int_{z_{1}}^{z_{2}} B_{0} sin(k z – ω t) h d z cos 0º \\ = \int_{z_{1}}^{z_{2}} B_{0} sin(k z - ω t) h d z \\ = \frac{- B_{0} h}{k} [cos(k z_{2} - ω t) – cos(k z_{1} – ω t)] ..... (2) \end{array}$
$\begin{array}{l} (iii) ϕ = \int B . d s = - \frac{B_{0} h}{k} [cos(k z_{2} - ω t] - \cos (k z_{1} - ω t)] \\ \frac{d ϕ}{d t} = - \frac{B_{0} h}{k} [– ω sin(k z_{2} - ω t) + ω sin(k z_{1} - ω t)] \\ = \frac{B_{0} h}{k} [ω sin(k z_{2} - ω t) – ω sin(k z_{1} - ω t)] \\ = \frac{B_{0} h ω}{k} [sin(k z_{2} - ω t) – sin(k z_{1} - ω t)] ..... (3) \\ Also, \oint E . d l = E_{0} h [sin(k z_{2} - ω t) – sin (k z_{1} - ω t)] ..... (4) \end{array}$
It is given that:
$\oint E . d l = \frac{d ϕ}{d t}$
From equation (3) and (4)
$\begin{array}{l} \Rightarrow E_{o} h [sin(k z_{2} – ω t) – sin(k z_{1} - ω t) \\ = \frac{B_{0} h ω}{k} [sin(k z_{2} - ω t) – sin(k z_{1} - ω t)] \\ \Rightarrow E_{0} h = \frac{B_{0} h ω}{k} \\ \Rightarrow E_{0} = (\frac{B_{0}}{k}) ω \\ \Rightarrow \frac{E_{0}}{B_{0}} = \frac{ω}{k} = c \end{array}$
(iv) Now consider the loop in y-z plane as shown in the diagram:

$\oint \vec{B} . d \vec{l} = \int_{1}^{2} \vec{B} . d \vec{l} + \int_{2}^{3} \vec{B} . d \vec{l} + \int_{3}^{4} \vec{B} . d \vec{l} + \int_{4}^{1} \vec{B} . d \vec{l}$
Here $\vec{B}$ = B₀sin(kz – $ω$ t) $\hat{j}$
Angle between $\vec{B}$ and side 1-2 is 0°.
Angle between $\vec{B}$ and side 2-3 is 90º.
Angle between $\vec{B}$ and side 3-4 is 180º
Angel between $\vec{B}$ and side 4-1 is 90º
So, $\oint B . d l = \int_{1}^{2} B d l cos 0º + 0 + \int_{3}^{4} B d l cos 180º + 0$
$\oint \vec{B} . d \vec{l} = B_{0} h [sin(k z_{1} - ω t) – sin(k z_{2} - ω t)]$
Now, flux due to electric field

$\begin{array}{l} ϕ_{E} = \int \vec{E} . d \vec{s} \\ = \int E d s cos 0 º {= \int}_{z_{1}}^{z_{2}} E_{0} sin(k z_{} - ω t) hdz \\ = \frac{- E_{0} h}{k} [cos(k z_{2} - ω t) – cos(k z_{1} - ω t)] \\ Now, \frac{d ϕ_{E}}{d t} = \frac{E_{0} h ω}{k} [sin(k z_{1} - ω t) – sin(k z_{2} - ω t)] \end{array}$
It is given that, conduction current,
$\begin{array}{l} \oint B . d l = μ_{0} (I+ ε_{0} \frac{d ϕ_{E}}{d t}) \\ \Rightarrow B_{0} h [sin(k z_{1} - ω t) – sin(k z_{2} - ω t)] \\ = 0 + μ_{0} ε_{0} \frac{E_{0} h ω}{k} [sin(k z_{1} - ω t) + sin(k z_{2} - ω t) \\ [For vacuum, I = 0] \\ \Rightarrow B_{0} = μ_{0} ε_{0} \frac{E_{0} ω}{k} \end{array}$
$\begin{array}{l} \Rightarrow \frac{E_{0}}{B_{0}} \times (\frac{ω}{k}) = \frac{1}{μ_{0} ε_{0}} \\ Here, \frac{E_{0}}{B_{0}} = \frac{ω}{k} = c \\ \Rightarrow c \times c = \frac{1}{μ_{0} ε_{0}} \\ \Rightarrow c^{2} = \frac{1}{μ_{0} ε_{0}} \\ \Rightarrow c = \frac{1}{\sqrt{μ_{0} ε_{0}}} \end{array}$

Page No 53:

Question 32:

A plane EM wave travelling along z direction is described by $E = E_{0} \sin (k z - ω t) \hat{i} and B = B_{0} \sin (k z - ω t) \hat{j}$ show that

(i) The average energy density of the wave is given by $u_{a v} = \frac{1}{4} ε_{0} {E^{2}}_{0} + \frac{1}{4} \frac{{B_{0}}^{2}}{4 μ_{0}}$
(ii) The time averaged intensity of the wave is given by $I_{a v} = \frac{1}{2} ε_{0} {E^{2}}_{0} .$

Answer:

(a) Energy contributed due to electric field,
$u_{E} = \frac{1}{2} ε_{0} E^{2}$
Energy contributed due to magnetic field,
$u_{(B)} = \frac{1}{2} \frac{B^{2}}{μ_{0}}$
So, total energy density:
u = u_(B)+ u_{(E)

$\begin{array}{l} = \frac{1}{2} ε_{0} E^{2} + \frac{1}{2} \frac{B^{2}}{μ_{0}} ..... (1) \\ The average value, \\ (E_{average}^{2}) = {[E_{0}^{2} {sin}^{2} (k z - ω t)]}_{average} \\ (B_{average}^{2}) = {[B_{0}^{2} {sin}^{2} (k z - ω t)]}_{average} \end{array}$}
By symmetry of graph of
$\sin^{2} θ (hard line) and \cos^{2} θ (dotted lin e) a s shown below :$

The average value of $\sin^{2} θ = average of \cos^{2} θ = \frac{1}{2}$
So, $E_{average}^{2} = \frac{E_{0}^{2}}{2} and B_{average}^{2} = \frac{B_{0}^{2}}{2}$

Putting the average values of
E² and B² in the equation (1),
We get:
$\begin{array}{l} u_{av} = \frac{1}{2} (ε_{0} \frac{E_{0}^{2}}{2}) = \frac{1}{2} \frac{(\frac{B_{0}^{2}}{2})}{μ_{0}} \\ = \frac{1}{4} ε_{0} E_{0}^{2} + \frac{1}{4} \frac{B_{0}^{2}}{μ_{0}} ..... (2) \end{array}$
(b) we have $\frac{E_{0}}{B_{0}} = c \Rightarrow B_{0} = \frac{E_{0}}{c}$
$\begin{array}{l} also c = \frac{1}{\sqrt{μ_{0} ε_{0}}} = c^{2} = \frac{1}{μ_{0} ε_{0}} \Rightarrow μ_{0} c^{2} = \frac{1}{ε_{0}} \\ So, from equation (2) \\ u_{av} = \frac{1}{4} ε_{0} E_{0}^{2} + \frac{1}{4} \times {(\frac{E_{0}}{c})}^{2} \times \frac{1}{μ_{0}} \\ = \frac{1}{4} ε_{0} E_{0}^{2} + \frac{1}{4} \times \frac{E_{0}^{2}}{c^{2} μ_{0}} = \frac{1}{4} ε_{0} E_{0}^{2} + \frac{1}{4} \frac{E_{0}^{2}}{(\frac{1}{ε_{0}})} \\ = \frac{1}{4} [ε_{0} E_{0}^{2} + ε_{0} E_{0}^{2}]= \frac{1}{2} ε_{0} E_{0}^{2} \\ So, I_{av} = \frac{1}{2} {cε}_{0} E_{0}^{2} = \frac{1}{2} I_{av} c \end{array}$

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